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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+//Book - Power system: Analysisi & Design 5th Edition
+//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.Overbye
+//Chapter-9 ;Example 9.6
+//Scilab Version - 6.0.0; OS - Windows
+
+clc;
+clear;
+
+
+Vf=1.05 //Prefault voltage in per unit
+Z0=%i*0.250 //Zero sequence impedance in per unit
+Z1=%i*0.13893 //Positive sequence impedance in per unit
+Z2=%i*0.14562 //Negative sequence impedance in per unit
+Zf=0 //Fault through impedance in per unit
+Zpr=0.20 //The positive sequence thevenin motor impedance at bus 2
+Zpl=0.455 //The positive sequence thevenin line impedance at bus 2
+Znr=0.21 //The negative sequence thevenin motor impedance at bus 2
+Znl=0.475 //The negative sequence thevenin line impedance at bus 2
+
+
+I1=Vf/(Z1+((Z0*Z2)/(Z0+Z2))) //Positive sequence fault current in per unit
+I2=-I1*(Z0/(Z0+Z2)) //Negative sequence fault current in per unit
+I0=-I1*(Z2/(Z0+Z2)) //Zero sequence fault current in per unit
+Iline0=0 //Zero sequence fault current from the line in per unit
+Imotor0=I0 //Zero sequence motor current from the motor in per unit
+Iline1=(Zpr/(Zpr+Zpl))*I1 //Positive sequence fault current from the line in per unit
+Ilead1=Iline1*exp(%i*(30)*(%pi/180)) //Positive sequence fault current from the line leads by 30 degree in per unit
+Imotor1=(Zpl/(Zpr+Zpl))*I1 //Positive sequence motor current from the motor in per unit
+Iline2=(Znr/(Znr+Znl))*I2 //Negative sequence fault current from the line in per unit
+Ilag2=Iline2*exp(%i*(-30)*(%pi/180)) //Negative sequence fault current from the line lags by 30 degree in per unit
+Imotor2=(Znl/(Znr+Znl))*I2 //Negative sequence motor current from the motor in per unit
+a=exp(%i*(120)*(%pi/180)) //operator a
+Iline=[1 1 1;1 (a^2) a;1 a (a^2)]*[0;Ilead1;Ilag2] //transforming the line currents to the phase domain
+Ilineb=Iline*0.41837 //transforming the line currents to the phase domain with base currents of 0.41837 kA
+
+disp(abs(clean(Iline,1e-10)),'The magnitude of transforming the line currents to the phase domain in per unit for each phase is given by:');
+disp(atand(imag(Iline),real(Iline)),'The angle of transforming the line currents to the phase domain in degreess for each phase is given by:');
+disp(abs(clean(Ilineb,1e-10)),'The magnitude of transforming the line currents to the phase domain in kA for each phase is given by:');
+disp(atand(imag(Ilineb),real(Ilineb)),'The angle of transforming the line currents to the phase domain in degreess for each phase is given by:');
+
+
+
+